Electrostatics Ques 177
- A conducting bubble of radius $a$, thickness $t(t \ll a)$ has potential $V$. Now the bubble collapses into a droplet. Find the potential of the droplet.
$(2005,2 \mathrm{M})$
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Answer:
Correct Answer: 177.$V^{\prime}=V \Big(\frac{a}{3t}\Big)^{1 / 3}$
Solution:
Formula:
Potential Difference Between Points A and B:
- Let $q$ be the charge on the bubble, then
$$ \begin{aligned} V & =\frac{K q}{a} & \text { Here, } K=\frac{1}{4 \pi \varepsilon_{0}} \\ \therefore \quad q & =\frac{V a}{K} & \end{aligned} $$
Let after collapsing, the radius of droplet becomes $R$, then equating the volume, we have
$$ \left(4 \pi a^{2}\right) t=\frac{4}{3} \pi R^{3} $$
$$ \therefore \quad R=\left(3 a^{2} t\right)^{1 / 3} $$
Now, potential of droplet will be $V^{\prime}=\frac{K q}{R}$
Substituting the values, we have
$$ V^{\prime}=\frac{(K) \frac{V a}{K}}{\left(3 a^{2} t\right)^{1 / 3}} \quad \text { or } \quad V^{\prime}=V \Big(\frac{a}{3t}\Big)^{1 / 3} $$
BETA
